Question: You have found the following ages (in years) of 4 tigers. The tigers are randomly selected from the 46 tigers at your local zoo: $ 11,\enspace 18,\enspace 8,\enspace 8$ Based on your sample, what is the average age of the tigers? What is the variance? You may round your answers to the nearest tenth.
Answer: Because we only have data for a small sample of the 46 tigers, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $4$ samples and divide by $4$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{0.09} + {44.89} + {10.89} + {10.89}} {{4 - 1}} $ $ {s^2} = \dfrac{{66.76}}{{3}} = {22.25\text{ years}^2} $ We can estimate that the average tiger at the zoo is 11.3 years old. There is a variance of 22.25 years $^2$.